E(x)??en?0(x?s),?(x)??2?Ven(x?s)2,s?(0)1/2 （3） 2?0en2、对于球坐标系，（1）为 （4）积分得
E(r)??1d2d?en （4） (r)??2rdrdr?0C1enen2C1r?2,?(r)??r??C2 3?0r6?0r带入边界条件?(R)??V,E(s)?0,?(s)?0，解得
ens3en22s3 E(r)??(r?2),?(r)??(r??3s2) （5）
3?0r6?0r鞘层厚度s满足
2s?3Rs?R?3、对于柱坐标系，（1）为 （7）积分得
E(r)??3236?0RV?0 （6） en1dd?en(r)?? （7） rdrdr?0Cenen2r?1,?(r)??r?C1lnr?C2 2?0r4?0带入边界条件?(R)??V,E(s)?0,?(s)?0，解得
ens2en22 E(r)??(r?),?(r)??(r?2s2lnr?s2?2slns) （8）
2?0r4?0鞘层厚度s满足
s?R?2slnR?2slns?
22224?0V?0 （9） en版权所有，违者必究！！
英文版低温等离子体作业
1-1、In a strictly steady state situation, both the ions and the electrons will follow the Boltzmann relation.
n?n0exp(?q?/kT)
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Show that the shielding distance is then given approximately by
?D?(?0kTeTi(Te?Ti)n0e2)1/2
and that
?D is determined by the temperature of the colder species.
解：英文版1.4节。
泊松方程满足 ????2e?0(ni?ne)??en0?0(exp(?e?/kTi)?exp(e?/kTe)) （1）
对（1）的右端做线性展开，保留电势的一阶项得 ???2en0e?(Te?Ti) （2）
?0kTeTi假设电势是球对称的，在球坐标系下（2）变成
1d2d?(Te?Ti)n0e2 2(r)?? （3）
rdrdr?0kTeT注意边界条件?|r???0,?|r?0??e/4??0r，解得电势分布并求出?D表达式 ???q4??0re?r/?D,?D?(?0kTeTi(Te?Ti)n0e2)1/2 （4）
当Te??Ti时，德拜长度
?D?(?0kTeTiTen0e2)1/2?(?0kTin0e2)1/2 （5）
取决于较小的温度Ti值。
2-1、The magnetic moment of a charged particle gyrating in a magnetic field is defined as the product of the current generated by the rotating particle times the area enclosed by the rotation. Show that this is equal to 证：粒子所受的力F满足
F?mv?/rc?qv?B?qv?B （1） 解得粒子回旋半径和回旋频率为
2??W?/B.
rc?mv?/qB?c?v?/rc?qB/m （2）
粒子在垂直磁场方向上圆周运动形成一个小的电流环，其电流满足
I?qf?q?c/2??qB/2?m （3）
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2所以，此电流环的磁矩为
2mv?2mv?Wq2B ??IS???()??? （4）
2?mqB2BB2-2、Consider a uniform magnetic field and a transverse electric field that varies slowly with time. Then the electric drift velocity also varies slowly with time. Therefore there is an inertial forceF??mdvDE/dt. Show that the polarization drift can be deduced by the expression of the drift in the general
force field. So it is also called inertial drift. 证：粒子在电场中的漂移速度为 vE?E?B （1） B2dvEmdE??2?B （2） dtBdt所以粒子在时变电场中所受的惯性力为 F??m粒子在一般力场中的漂移速度为 vF?F?B （3） qB2将（2）代入（3），注意E?B，得 vDP??mdEmdEdE(?B)?B??(B(?B)?(B?B)) 44qBdtqBdtdt ?mdE （4）
qB2dt这正是极化漂移的速度公式。
2-3、Consider the magnetic mirror system with length L. The magnetic field may be approximated by B(z)?B0(1?3z2/L2),where denotes the coordinate from the midplane along the field. z (1) which particle will be confined? (2) Calculate the probability of loss.
(3) Show that particle motion is simple harmonic and give out the frequency. 解：1、1、由B(z)分布，可以求出Bm?4B0，由磁矩守恒得
1212mv0?mvm?122? ，即v0??vm? （1） 2B0Bm当粒子能被约束时，由粒子能量守恒有vm??v0，因此带电粒子能被约束住的条件是在磁镜
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中央，粒子速度满足v0??2、逃逸粒子百分比
1v0 22?1 P?2???d??sin?d??1?003?13.4% （2） 23、在z轴方向，粒子受力F等于 Fz???粒子运动方程为
6?BdBz??20z （3） dzL???Fz?? mz6?B0z （4） 2L粒子运动为简谐振动，其频率为??6?B0/L。
5-1、Assuming that the distribution function for electrons is the Druyvesteyn distribution, calculate the average electron energy and the directed velocity. 解：德留维斯坦分布为
23me? f?Aexp(?222v4) （1）
8eE?ea归一化系数A满足
23me? 1??f(v)dv?4?A?exp(?222v4)v2dv （2）
8eE?ea03?3me2?4令??222v代入(2)得
8eE?ea23me? ?A(222)?3/4?exp(??)??1/4d??1 （3）
8eE?ea02?133/4me0.37me?3/2所以归一化系数A?()(222)3/4?()。
??(3/4)8eE?ea?eE?ea23me?123平均动能为Ke??mevfdv?2?meA?exp(?222v4)v4dv
28eE?ea022me3me?3/43me2??5/4 ?(222)(222)?exp(??)?1/4d?
?(3/4)8eE?ea8eE?ea02?2me/ma?1/2?(5/4)3me3?(5/4) ?2me()?ma/meeE?ea 2?(3/4)8e2E2?ea3?(3/4)??? ?0.4ma/meeE?ea
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4?eE?eadf24?eE?ea定向速度为ue?vdv??3?medv3me??0?fdv2
223me?3/43me?4?eE?ea1 ??(222)?exp(?222v4)dv2
3me??(3/4)8eE?ea8eE?ea02eE?ea3me?3/4me2??1/24 ??(222)(222)?exp(??2)d?
3?(3/4)me8eE?eaeE?ea0? ??2?31/41/4eE?ea1/2()?()?0.69?1/4(eE?ea/me)1/2
3?(3/4)8me*7-1、Consider a high-pressure steady-state discharge confined inside of a rectangular box having edges of length a meters along x, b meters along y, and c meters along z. The center of the box is located at x?0,y?0,z?0. The plasma is created by a volume ionization G??ineand is lost to the walls by ambipolar diffusion with a constant ambipolar diffusion coefficient Da. Here iν is the electron-neutral ionization frequency. Assume that the electron density
ne is n0 in the center of the box and is zero on the walls.
(a) Find an expression for the density ne(x,y,z) inside the box. (b) Find the relation between Da,?iand the dimensions of the box. 解：由平衡态粒子数守恒方程得?Da?2n??ine，化简得亥姆霍兹方程 ?2ne?k2ne?0,k??i/Da （1）
对（1）分离变量法求解。设ne?X(x)Y(y)Z(z)，有
??2X??x2X?0 (2.1)?22??Y??yY?0 (2.2) ?2 2??Z??zZ?0 (2.3)??2??2??2?k2 (2.4)yz?x解方程（2.1），考虑到边界条件X(?a/2)?X(a/2)?0和X?0得
X(x)?cos?ax,?x??a （3.1）
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